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Karo-lina-s [1.5K]
2 years ago
12

One of the zeros of the polynomial function is 5.

Mathematics
2 answers:
maks197457 [2]2 years ago
7 0
The last term  is -100 so  its either a or d.
Try a:-

(x^2 - 25)(x-2)(x - 2)

= x^2 - 25 (x^2 - 4x + 4)

= x^2 + 4x^3 + 4x^2 - 25x^2 + 100x - 100
= x^4 + 4x^3 - 21x^2 + 100x - 100   

Its a.

astra-53 [7]2 years ago
7 0

Answer:

The factored form of the function is:

A. f(x) = (x-5)(x+5)(x-2)^{2}

Step-by-step explanation:

To factor the expression you can use synthetic division.

f(x)=x4−4x3−21x2+100x−100

1. Take the constant and find it's factors. So the factors of 100 are:     ±1, ±2, ±4, ±5,±10,±20,±25, ±50,±100

  • In this case -5 works (synthetic division is shown in the picture attached)

<em>        Hint: you know a number divides the polynomial expression      </em>

<em>        when the remainder is zero </em>

2. Therefore: f(x) = (x+5)(x^{3}-9x^{2}+24x-20)

   Using synthetic division once more. The factors of twenty are:

   ±1, ±2, ±4, ±5,±10,±20.

   In this case 2 works (synthetic division is shown in the picture          

   attached)

3. Therefore: f(x)= (x+5)(x-2)(x^{2} -7x+10)

4. Factor the last expression: (x^{2} -7x+10)

  • _ + _ = -7 and _ x _= 10
  • -5 and -2

5. So (x^{2} -7x+10) = (x-5)(x-2)

6. Therefore: f(x) = (x+5)(x-2)(x-5)(x-2) = (x-5)(x+5)(x-2)^{2}

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Arranging data in ascending order to identify the outlier and calculate the mean

Step-by-step explanation:

Given the data as: [4, 1, 3, 10, 18, 12, 9, 4, 15, 16, 32]

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The outlier is 32

Finding the mean;

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Which of the following represents the zeros of f(x) = 5x3 − 6x2 − 59x + 12? (2 points)
Furkat [3]

Answer:

4, −3, 1 over 5

Step-by-step explanation:

x is a zero of f(x) if f(x) = 0

In this problem, we have that:

f(x) = 5x^{3} - 6x^{2} - 59x + 12

4, 3, 1 over 5

f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0

So 4 is a zero of the function

f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84

So 3 is not a zero of the function, and this option is incorrect

4, 3, − 1 over 5

f(3) = 5*3^{3} - 6*3^{2} - 59*3 + 12 = -84

So 3 is not a zero of the function, and this option is incorrect

4, −3, 1 over 5

f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0

So 4 is a zero of the function

f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0

So -3 is a zero of the function

f(\frac{1}{5}) = f(0.2) = 5*(0.2)^{3} - 6*(0.2)^{2} - 59*(0.2) + 12 = 0

So 1 over 5 is a zero of the function

This is the correct answer.

4, −3, −1 over 5

f(4) = 5*4^{3} - 6*4^{2} - 59*4 + 12 = 0

So 4 is a zero of the function

f(-3) = 5*(-3)^{3} - 6*(-3)^{2} - 59*(-3) + 12 = 0

So -3 is a zero of the function

f(-\frac{1}{5}) = f(-0.2) = 5*(-0.2)^{3} - 6*(-0.2)^{2} - 59*(-0.2) + 12 = 23.52

-1 over 5 is not a zero of the function

5 0
2 years ago
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